Minimal Polynomial

Cayley-Hamilton Theorem

Every square matrix satisfies its own characteristic equation.
An+d1An−1+⋯+dn−1A+dnI=0 \mathbf{A}^n + d_1\mathbf{A}^{n-1} + \cdots + d_{n-1}\mathbf{A} + d_n\mathbf{I} = 0

Remark: Cayley-Hamilton Theorem basically states that nth and higher powers of a matrix can be expressed in terms of lower powers of the matrix. This is a very useful property in solving linear systems.

An=−d1An−1−⋯−dn−1A−dnI A^n = -d_1A^{n-1} - \cdots - d_{n-1}A - d_nI

For previous example, we have:

A2−4A+3I=0 A^2 - 4A + 3I = 0

A2=4A−3I A^2 = 4A - 3I

A3=4A2−3A=4(4A−3I)−3A=13A−12I A^3 = 4A^2 - 3A = 4(4A - 3I) - 3A = 13A - 12I

A4=4A3−3A2=4(13A−12I)−3(4A−3I)=49A−48I A^4 = 4A^3 - 3A^2 = 4(13A - 12I) - 3(4A - 3I) = 49A - 48I


exp(s)=∑n=0∞snn!=I+s+s22!+s33!+⋯ \large exp(s) = \sum_{n=0}^{\infty} \frac{s^n}{n!} = I + s + \frac{s^2}{2!} + \frac{s^3}{3!} + \cdots

exp(A)=∑n=0∞Ann!=I+A+A22!+A33!+⋯ \large exp(A) = \sum_{n=0}^{\infty} \frac{A^n}{n!} = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots


A−1=?A^{-1} = ?

A2−4A+3I=0A^2 - 4A + 3I = 0

A−4I+3A−1=0A - 4I + 3A^{-1} = 0

A−1=(4I−A)3A^{-1} = \Large \frac{(4I - A)}{3}


Example: A=[2112]A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} A100=?A^{100}=?

Solution:

A100=αA+βIA^{100} = \alpha A + \beta I

λ1=3,λ2=1, and e1=[11],e2=[1−1]\lambda_1 = 3, \lambda_2 = 1 \text{, and } e_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, e_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}

Ae1=λ1e1⇒A2e1=λ12e1⇒A3e1=λ13e1⇒⋯⇒A100e1=λ1100e1 A e_1 = \lambda_1 e_1 \Rightarrow A^2 e_1 = \lambda_1^2 e_1 \Rightarrow A^3 e_1 = \lambda_1^3 e_1 \Rightarrow \cdots \Rightarrow A^{100} e_1 = \lambda_1^{100} e_1

Ae2=λ2e2⇒A2e2=λ22e2⇒A3e2=λ23e2⇒⋯⇒A100e2=λ2100e2 A e_2 = \lambda_2 e_2 \Rightarrow A^2 e_2 = \lambda_2^2 e_2 \Rightarrow A^3 e_2 = \lambda_2^3 e_2 \Rightarrow \cdots \Rightarrow A^{100} e_2 = \lambda_2^{100} e_2

A100=αA+βI⇒A100e1=αAe1+βe1⇒λ1100e1=αλ1e1+βe1⇒λ1100=αλ1+β  ⟹  3100=3α+β A^{100} = \alpha A + \beta I \Rightarrow A^{100} e_1 = \alpha A e_1 + \beta e_1 \Rightarrow \lambda_1^{100} e_1 = \alpha \lambda_1 e_1 + \beta e_1 \Rightarrow \lambda_1^{100} = \alpha \lambda_1 + \beta \implies 3^{100} = 3\alpha + \beta

A100=αA+βI⇒A100e2=αAe2+βe2⇒λ2100e2=αλ2e2+βe2⇒λ2100=αλ2+β  ⟹  1100=α+β A^{100} = \alpha A + \beta I \Rightarrow A^{100} e_2 = \alpha A e_2 + \beta e_2 \Rightarrow \lambda_2^{100} e_2 = \alpha \lambda_2 e_2 + \beta e_2 \Rightarrow \lambda_2^{100} = \alpha \lambda_2 + \beta \implies 1^{100} = \alpha + \beta

{3100=3α+β1=α+β⇒α=3100−12β=−3100+32 \begin{cases} 3^{100} = 3\alpha + \beta \\ 1 = \alpha + \beta \end{cases} \Rightarrow \begin{matrix} \alpha = \frac{3^{100} - 1}{2} \\ \beta = \frac{-3^{100} + 3}{2} \end{matrix}

A100=3100−12A+−3100+32I A^{100} = \frac{3^{100} - 1}{2} A + \frac{-3^{100} + 3}{2} I


Minimal Polynomial

Definition: The monic polynomial is the polynomial with the highest degree coefficient equal to 1.

sn+a1sn−1+⋯→ s^n + a_1s^{n-1} + \cdots \rightarrow monic

2sn+a1sn−1+⋯→ 2s^n + a_1s^{n-1} + \cdots \rightarrow not monic

Definition: The minimal polynomial of a matrix is the monic polynomial of the lowest degree that has the matrix as a root.

m(A)=0n×nm(A) = 0_{n \times n}

Theorem: Given a matrix A=Cn timesnA = \mathbb{C}^{n \ times n}, let m(s) be its minimal polynomial. Then, the minimal polynomial is the smallest degree polynomial that satisfies the following:

i) m(s) is unique.
ii) m(s) divides d(s) with no remainder.
∃q(s) s.t. d(s)=q(s)m(s) \exists q(s) \text{ s.t. } d(s) = q(s)m(s)
iii) Every root of m(s) is a root of d(s).

Example: A=[100020003]A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} d(s)=(s−1)(s−2)(s−3)m(s)=(s−1)(s−2)(s−3)\Large \substack{d(s) = (s-1)(s-2)(s-3) \\ m(s) = (s-1)(s-2)(s-3)}

Remark: When A has distinct eigenvalues (which implies diagonalizability but converse is not true), the minimal polynomial is the same as the characteristic polynomial.


Example: A1=[100020002]A_1 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} d(s)=(s−1)(s−2)2m(s)=(s−1)(s−2)       A2=[100021002]\Large \substack{d(s) = (s-1)(s-2)^2 \\ m(s) = (s-1)(s-2)} \ \ \ \ \ \ \ \normalsize A_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} d(s)=(s−1)(s−2)2m(s)=(s−1)(s−2)2\Large \substack{d(s) = (s-1)(s-2)^2 \\ m(s) = (s-1)(s-2)^2}

A way to check if a minimal polynomial is correct is to check if the characteristic polynomial is zero when the minimal polynomial is substituted for s.

Let m1(s)=(s−1)(s−2)m_1(s) = (s-1)(s-2) then m1(A)=(A−1I)(A−2I)=[000010001][−100000000]=[000000000]✓m_1(A) = (A-1I)(A-2I) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \checkmark

Let m2(s)=(s−1)(s−2)m_2(s) = (s-1)(s-2) then m2(A)=(A−1I)(A−2I)=[000010001][−100001000]=[000001000]≠0m_2(A) = (A-1I)(A-2I) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \neq 0

Then, m2(s)=(s−1)(s−2)2  ⟹  m2(A)=(A−1I)(A−2I)2=[000010001][−100001000][−100001000]=[000000000]✓m_2(s) = (s-1)(s-2)^2 \implies m_2(A) = (A-1I)(A-2I)^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \checkmark

R3=N(A−1I)⊕N((A−2I)2) \mathbb{R}^3 = N(A-1I) \oplus N((A-2I)^2) where,

N(A−1I)=N([000010001])=span{[100]}  dim(N(A−1I))=1 N(A-1I) = N \bigg ( \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \bigg ) = \text{span} \bigg \{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \bigg \} \ \ \text{dim} \bigg ( N(A-1I) \bigg ) = 1

N((A−2I)2)=N([100000000])=span{[010],[001]}  dim(N((A−2I)2))=2 N((A-2I)^2) = N \bigg ( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \bigg ) = \text{span} \bigg \{ \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg \} \ \ \text{dim} \bigg ( N((A-2I)^2) \bigg ) = 2

Remark: There is a relationship between decomposition form and minimal polynomial in terms of power.

m(s)=(s−λ1)d1⋯(s−λk)dk m(s) = (s-\lambda_1)^{d_1} \cdots (s-\lambda_k)^{d_k}
Rn=N(A−λ1I)d1⊕⋯⊕N(A−λkI)dk \mathbb{R}^n = N(A-\lambda_1I)^{d_1} \oplus \cdots \oplus N(A-\lambda_kI)^{d_k}

Theorem:

Cn=N(A−λ1I)m1⊕N(A−λ2I)m2⊕⋯⊕N(A−λkI)mk\mathbb{C}^n = N(A-\lambda_1I)^{m_1} \oplus N(A-\lambda_2I)^{m_2} \oplus \cdots \oplus N(A-\lambda_kI)^{m_k}
d(s)=(s−λ1)r1⋯(s−λk)rkd(s) = (s-\lambda_1)^{r_1} \cdots (s-\lambda_k)^{r_k}
r1+r2+⋯+rk=nr_1 + r_2 + \cdots + r_k = n
m(s)=(s−λ1)m1⋯(s−λk)mkm(s) = (s-\lambda_1)^{m_1} \cdots (s-\lambda_k)^{m_k}
1≤mi≤ri1 \leq m_i \leq r_i

Aˉ=[Aˉ10⋯00Aˉ2⋯0⋮⋮⋱⋮00⋯Aˉk]\bar A = \begin{bmatrix} \bar A_1 & 0 & \cdots & 0 \\ 0 & \bar A_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \bar A_k \end{bmatrix} Aˉ=B−1AB, B is composed of the basis vectorsB=[e1e2⋯en] for the N(A−λiI)mi\Large \substack{ \bar A = B^{-1}AB \text{, B is composed of the basis vectors} \\ B = \begin{bmatrix} e_1 & e_2 & \cdots & e_n \end{bmatrix} \text{ for the } N(A-\lambda_iI)^{m_i} }

Size of Aˉi\bar A_i is dim(N(A−λiI)mi)\text{dim} \bigg ( N(A-\lambda_iI)^{m_i} \bigg )


Example: A=[1000011000100002]A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix} d(s)=(s−1)3(s−2)m(s)=(s−1)2(s−2)\Large \substack{d(s) = (s-1)^3(s-2) \\ m(s) = (s-1)^2(s-2)}

Solution: Let Σ1\Sigma_1 be A−λ1IA-\lambda_1I and Σ2\Sigma_2 be A−λ2IA-\lambda_2I.

Σ1=[0000001000000001]dim(N(Σ1))=2≠3=r1→dimV−dimR(Σ1)=4−2=2\Sigma_1 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1) \bigg ) = 2 \neq 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1)}= 4 - 2 = 2

Then we need to check the dimension of N(Σ12)N(\Sigma_1^2).

Σ12=[0000001000000001][0000001000000001]=[0000000000000001]dim(N(Σ12))=1=r1→dimV−dimR(Σ12)=4−1=3=r1\Sigma_1^2 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^2) \bigg ) = 1 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^2)} = 4 - 1 = 3 = r_1

Lets briefly check the dimension of N(Σ13)N(\Sigma_1^3).

Σ13=[0000000000000001]dim(N(Σ13))=3=r1→dimV−dimR(Σ13)=4−1=3=r1\Sigma_1^3 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^3) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^3)} = 4 - 1 = 3 = r_1

                      ⋮ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots

Σ14=[0000000000000001]dim(N(Σ14))=3=r1→dimV−dimR(Σ14)=4−1=3=r1\Sigma_1^4 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^4) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^4)} = 4 - 1 = 3 = r_1

Lets check the second eigenvalue.

Σ2=[−10000−11000−100000]=[100001−1000100000]dim(N(Σ2))=1=r2→dimV−dimR(Σ2)=4−3=1\Sigma_2 = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2) \bigg ) = 1 = r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2)} = 4 - 3 = 1

Then the minimal polynomial is m(s)=(s−1)2(s−2)m(s) = (s-1)^2(s-2).


Example: A=[1100011000100002]A = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix} d(s)=(s−1)3(s−2)m(s)=(s−1)3(s−2)\Large \substack{d(s) = (s-1)^3(s-2) \\ m(s) = (s-1)^3(s-2)}

Solution: Let Σ1\Sigma_1 be A−λ1IA-\lambda_1I and Σ2\Sigma_2 be A−λ2IA-\lambda_2I.

Σ1=[0100001000000001]dim(N(Σ1))=1≠r1→dimV−dimR(Σ1)=4−3=1\Sigma_1 = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1) \bigg ) = 1 \neq r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1)} = 4 - 3 = 1

Then we need to check the dimension of N(Σ12)N(\Sigma_1^2).

Σ12=[0010000000000001]dim(N(Σ12))=2≠r1→dimV−dimR(Σ12)=4−2=2\Sigma_1^2 = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^2) \bigg ) = 2 \neq r_1 \rightarrow \text{dim}{V}- \text{dim}{R(\Sigma_1^2)} = 4 - 2 = 2

Check the dimension of N(Σ13)N(\Sigma_1^3).

Σ13=[0000000000000001]dim(N(Σ13))=3=r1→dimV−dimR(Σ13)=4−1=3\Sigma_1^3 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^3) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^3)}= 4 - 1 = 3

Now we can check the second eigenvalue.

Σ2=[−11000−11000−100000]=[1−10001−1000100000]dim(N(Σ2))=1=r2→dimV−dimR(Σ2)=4−3=1\Sigma_2 = \begin{bmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2) \bigg ) = 1 = r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2)} = 4 - 3 = 1

Then the minimal polynomial is m(s)=(s−1)3(s−2)m(s) = (s-1)^3(s-2).


Example: A=[2100000000021000000000200000000002100000000020000000000200000000002000000000031000000000300000000003]A = \begin{bmatrix} 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix} d(s)=(s−2)7(s−3)3m(s)=?\Large \substack{d(s) = (s-2)^7(s-3)^3 \\ m(s) = ?}

Solution: For each jordan block, we need to check the dimension of N(Σ1i)N(\Sigma_1^i).

The largest jordan block will have the largest dimension of N(Σ1i)N(\Sigma_1^i). The rest of the jordan blocks will have dimension of N(Σ1i)N(\Sigma_1^i) equal to the size of the jordan block. Hence the geometric multiplicity wont increase.

Let Σ1\Sigma_1 be A−λ1IA-\lambda_1I and Σ2\Sigma_2 be A−λ2IA-\lambda_2I. Check the largest jordan block.

Σ1=[010001000]dim(N(Σ1))=1≠r1→dimV−dimR(Σ1)=3−2=1\Sigma_1 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1) \bigg ) = 1 \neq r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1)} = 3 - 2 = 1

Then we need to check the dimension of N(Σ12)N(\Sigma_1^2).

Σ12=[001000000]dim(N(Σ12))=2≠r1→dimV−dimR(Σ12)=3−1=2\Sigma_1^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^2) \bigg ) = 2 \neq r_1 \rightarrow \text{dim}{V}- \text{dim}{R(\Sigma_1^2)} = 3 - 1 = 2

Check the dimension of N(Σ13)N(\Sigma_1^3).

Σ13=[000000000]dim(N(Σ13))=3=r1→dimV−dimR(Σ13)=3−0=3\Sigma_1^3 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^3) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^3)} = 3 - 0 = 3

We need to continue to see that further powers of Σ1\Sigma_1 will not increase the dimension of N(Σ1i)N(\Sigma_1^i).

Σ14=[000000000]dim(N(Σ14))=3=r1→dimV−dimR(Σ14)=3−0=3\Sigma_1^4 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^4) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^4)} = 3 - 0 = 3

This implies that the further jordan blocks will not increase the dimension of N(Σ1i)N(\Sigma_1^i). Hence they are not needed to be checked for minimal polynomial.

Now we can check the second eigenvalue.

Σ2=[0−200]dim(N(Σ2))=1≠r2→dimV−dimR(Σ2)=2−1=1\Sigma_2 = \begin{bmatrix} 0 & -2 \\ 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2) \bigg ) = 1 \neq r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2)} = 2 - 1 = 1

Σ22=[0000]dim(N(Σ22))=2=r2→dimV−dimR(Σ22)=2−0=2\Sigma_2^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2^2) \bigg ) = 2 = r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2^2)} = 2 - 0 = 2

Then the minimal polynomial is m(s)=(s−2)7(s−3)2m(s) = (s-2)^7(s-3)^2.

Remark:

dim(N(A−λiI))=# of jordan blocks corresponding to λi with size ≥ 1\text{dim} \bigg ( N(A-\lambda_iI) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ 1}
dim(N(A−λiI)2)−dim(N(A−λiI))=# of jordan blocks corresponding to λi with size ≥ 2\text{dim} \bigg ( N(A-\lambda_iI)^2 \bigg ) - \text{dim} \bigg ( N(A-\lambda_iI) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ 2}
dim(N(A−λiI)3)−dim(N(A−λiI)2)=# of jordan blocks corresponding to λi with size ≥ 3\text{dim} \bigg ( N(A-\lambda_iI)^3 \bigg ) - \text{dim} \bigg ( N(A-\lambda_iI)^2 \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ 3}
dim(N(A−λiI)k)−dim(N(A−λiI)k−1)=# of jordan blocks corresponding to λi with size ≥ k\text{dim} \bigg ( N(A-\lambda_iI)^k \bigg ) - \text{dim} \bigg ( N(A-\lambda_iI)^{k-1} \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ k}

Example: A∈R4A \in \mathbb{R}^4 Which has a single eigenvalue λ1=7\lambda_1 = 7 and the geometric multiplicity is 2. Find all possible jordan forms.

Solution: The characteristic polynomial is d(s)=(s−7)4d(s) = (s-7)^4 and the minimal polynomial is m(s)=(s−7)2m(s) = (s-7)^2. This suggests that,

dim(N(A−7I)2)=4=r1→dim(V)=dim(N(A−7I)2)\text{dim} \bigg ( N(A-7I)^2 \bigg ) = 4 = r_1 \rightarrow \text{dim}(V) = \text{dim} \bigg ( N(A-7I)^2 \bigg )

Then for the dim(N(A−7I))\text{dim}\bigg (N(A-7I)\bigg ) we have 3 possibilities, 1, 2, or 3.

a) Let dimN(A−7I)=3 \text{a) Let } \text{dim} N(A-7I) = 3

     # of jordan blocks =3 \ \ \ \ \ \text{\# of jordan blocks } = 3

     # of jordan blocks w/ size ≥2=4−3=1 \ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 2 = 4 - 3 = 1

     # of jordan blocks w/ size ≥3=3−3=0 \ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 3 = 3 - 3 = 0

A=[7100070000700007]A = \begin{bmatrix} 7 & 1 & 0 & 0 \\ 0 & 7 & 0 & 0 \\ 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & 7 \end{bmatrix}

a) Let dimN(A−7I)=2 \text{a) Let } \text{dim} N(A-7I) = 2

     # of jordan blocks =2 \ \ \ \ \ \text{\# of jordan blocks } = 2

     # of jordan blocks w/ size ≥2=4−2=2 \ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 2 = 4 - 2 = 2

     # of jordan blocks w/ size ≥3=dim(N(A−7I)3)− dim(N(A−7I)2)=4−4=0 \ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 3 = \text{dim}\bigg ( N(A-7I)^3 \bigg ) - \text{ dim}\bigg ( N(A-7I)^2 \bigg ) = 4 - 4 = 0

A=[7100070000710007]A = \begin{bmatrix} 7 & 1 & 0 & 0 \\ 0 & 7 & 0 & 0 \\ 0 & 0 & 7 & 1 \\ 0 & 0 & 0 & 7 \end{bmatrix}

a) Let dimN(A−7I)=1 \text{a) Let } \text{dim} N(A-7I) = 1

     # of jordan blocks =1 \ \ \ \ \ \text{\# of jordan blocks } = 1

     # of jordan blocks w/ size ≥2=4−1=3 \ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 2 = 4 - 1 = 3

However, this is not possible since there cannot be 3 jordan blocks with size greater than or equal to 2.


Example: Suppose a matrix A∈R8×8A \in \mathbb{R}^{8\times 8} has the following subspace dimensions:

dim(N(A−3I))=5dim(N(A−3I)2)=7dim(N(A−3I)3)=8 \text{dim} \bigg ( N(A-3I) \bigg ) = 5 \\ \text{dim} \bigg ( N(A-3I)^2 \bigg ) = 7 \\ \text{dim} \bigg ( N(A-3I)^3 \bigg ) = 8

a) Find the characteristic polynomial of A.

b) Find the minimal polynomial of A.

c) Find the possible Jordan forms of A.

Solution:

a) We know that the characteristic polynomial is d(s)=(s−3)8d(s) = (s-3)^8.

b) We know that the minimal polynomial is m(s)=(s−3)3m(s) = (s-3)^3 since the dimension of N(A−3I)3N(A-3I)^3 is same as the dimension of the vector space. This implies that the geometric multiplicity is 3.

c) Starting from the largest jordan block, having geometric multiplicity 3 suggest that the largest jordan block has size 3.

dim(N(A−3I)3)−dim(N(A−3I)2)=# of jordan blocks corresponding to λ1 with size ≥3=8−7=1 \text{dim} \bigg ( N(A-3I)^3 \bigg ) - \text{dim} \bigg ( N(A-3I)^2 \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_1 \text{ with size }\geq 3 = 8 - 7 = 1

dim(N(A−3I)2)−dim(N(A−3I))=# of jordan blocks corresponding to λ1 with size ≥2=7−5=2 \text{dim} \bigg ( N(A-3I)^2 \bigg ) - \text{dim} \bigg ( N(A-3I) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_1 \text{ with size }\geq 2 = 7 - 5 = 2

We have 2 that are greater than or equal to 2. This implies that we have 1 jordan block with size 3 and 1 jordan blocks with size 2.

dim(N(A−3I))=# of jordan blocks corresponding to λ1 with size ≥1=5 \text{dim} \bigg ( N(A-3I) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_1 \text{ with size }\geq 1 = 5

This implies that we have 3 jordan blocks with size 1.

A=[3100000003100000003000000003100000003100000003000000003100000003]A = \begin{bmatrix}3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix}


#EE501 - Linear Systems Theory at METU